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Prove that if w, z are complex numbers such that |w| = |z| = 1 and \(wz\ne-1\), then \(\frac{w+z}{1+wz}\) is a real number.

 Sep 19, 2018
 #1
avatar+21358 
+9

Prove that if w, z are complex numbers such that \(|w| = |z| = 1\) and \(wz\ne-1\),

then \(\dfrac{w+z}{1+wz}\)  is a real number.

 

\(\begin{array}{|rcll|} \hline |w| = 1 ~&\Rightarrow& ~|w|^2 = 1 \\ |z| = 1 ~&\Rightarrow& ~|z|^2 = 1 \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline && \dfrac{w+z}{1+wz} \\\\ &=& \dfrac{w}{1+wz} + \dfrac{z}{1+wz} \\\\ &=& \dfrac{1}{ \dfrac{1}{w} + z} + \dfrac{1} { \dfrac{1}{z} + w} \quad | \quad \dfrac{1}{w} = \dfrac{\bar{w}}{|w|^2}=\bar{w},\ \quad \dfrac{1}{z} = \dfrac{\bar{z}}{|z|^2}=\bar{z} \\\\ &=& \dfrac{1}{ \bar{w} + z} + \dfrac{1} { \bar{z} + w} \\\\ &=& \dfrac{ \bar{z} + w+\bar{w} + z}{(\bar{w} + z)(\bar{z} + w)} \\\\ &=& \dfrac{ ( w+\bar{w}) + (z+\bar{z} )}{(\bar{w} + z)(w+\bar{z})} \quad | \quad w = a+bi,\quad \bar{w} = a-bi, \quad z=c+di, \quad \bar{z}=c-di \\\\ &=& \dfrac{ ( a-bi+a-bi) + (c+di+c-di )}{(a-bi + c+di)(a+bi+c-di)} \\\\ &=& \dfrac{ 2( a+c) }{[~(a+c)+(d-b)i~][~(a+c)-(d-b)i~]} \\\\ &=& \dfrac{ 2( a+c) }{[~(a+c)^2-(d-b)^2i^2~]} \quad | \quad i^2=-1 \\\\ &=& \dfrac{ 2( a+c) }{ (a+c)^2+(d-b)^2 } \\ \hline \end{array}\)

 

laugh

 Sep 20, 2018
 #2
avatar+4032 
+2

\(\dfrac{w+z}{1+w z}= \dfrac{(w+z)(1+w z)^*}{(1+w z)(1+w z)^*} = \\ \dfrac{w+z + w(wz)^* + z(wz)^*}{|1+wz|^2}\)

 

The denominator is real so we just need to show the numerator is also real

 

\(\text{Noting that }(wz)^* = w^* z^* \text{ we have}\\ w+z+w(wz)^*+z(wz)^* = \\ w + z + w w^* z^* + z z^* w^* = \\ w + z + 1\cdot z^* + 1 \cdot w^* = \\ (w+w^*)+(z+z^*) = 2Re(w) + 2Re(z) = \\ 2Re(w+z) \in \mathbb{R} \)

.
 Sep 20, 2018

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